Calculate the ratio by using Henderson-Hasselbalch equation:
pH = pKa + log [neutral form] / Protonated form
3.05 = 2.21 + log [neutral form] / [Protonated form]
3.05 - 2.21 = log [neutral form] / [Protonated form]
0.84 = log [neutral form] / [Protonated form]
[neutral form] / [protonated form] = anti log 0.84 = 6.91
3 ..............................
Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ
Potassium itself is not a compound, it’s an element. Represented by “K” and has an atomic number of 19. However, it can be used to make a compound
